Question

Two long straight wires are arranged parallel to each other and are kept \(20\) cm apart in vacuum. They carry currents of \(2\) A and \(4\) A respectively in the same direction. What will be the magnetic force on a length of \(10\) cm of either wire?
(\(μ_0 = 4π\times 10^{-7}\) SI units)

• A. \(10^{-7}\) N
• B. \(2\times 10^{-7}\) N
• C. \(4\times 10^{-7}\) N
• D. \(8\times 10^{-7}\) N

Hint:

Force on length \(l\) is \(F=\frac{\mu_0I_1I_2l}{2\pi d}\).

Answer 1

The Correct Option is D

Step 1: Key Formula:
Force between two long parallel wires: \(F = \frac{\mu_0}{2\pi}\frac{I_1I_2}{d}l\). It is attractive for currents in the same direction.

Step 2: Calculate:
\(\frac{\mu_0}{2\pi} = 2\times10^{-7}\) SI units. \(I_1 = 2\) A, \(I_2 = 4\) A, \(d = 0.2\) m, \(l = 0.1\) m.
\[ F = 2\times10^{-7}\times\frac{2\times4}{0.2}\times0.1 = 2\times10^{-7}\times4 = 8\times10^{-7}\ \text{N} \]

Final Answer:
The force is \(8\times10^{-7}\) N, option (D). \[ \boxed{8\times10^{-7}\ \text{N}} \]